Question

Prove the following question​

Answer 1

Answer:

57788+47885+379-6799+3589

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Prove the following :

$\sf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} + {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2} = 1$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\red{\rm :\longmapsto\:\sf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} + {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2}}$

Now, Consider

$\red{{\rm :\longmapsto\:\sf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} }}$

We know,

$\boxed{ \bf{ \: {x}^{m} \times {y}^{m} = {(xy)}^{m} }}$

So, using this identity, we get

$\sf \: = \: { \bigg(cosx(3 - 4 {cos}^{2} x) \bigg)}^{2}$

$\sf \: = \: { \bigg(3cosx - 4 {cos}^{3} x \bigg)}^{2}$

$\sf \: = \: { \bigg( - (4 {cos}^{3} x - 3cosx) \bigg)}^{2}$

We know that,

$\boxed{ \bf{ \: cos3x = {4cos}^{3}x - 3cosx}}$

So, using this identity, we get

$\sf \: = \: {( - cos3x)}^{2}$

$\sf\: = \: {cos}^{2}3x$

Hence,

${\rm :\longmapsto\:\bf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} } = \bf{cos}^{2}3x$

Now, Consider

$\red{\rm :\longmapsto\:\sf \: {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2}}$

We know,

$\boxed{ \bf{ \: {x}^{m} \times {y}^{m} = {(xy)}^{m} }}$

So, using this, we get

$\sf \: = \: { \bigg(sinx(3 - 4 {sin}^{2} x) \bigg)}^{2}$

$\sf \: = \: { \bigg(3sinx - 4 {sin}^{3} x \bigg)}^{2}$

We know,

$\boxed{ \bf{ \: sin3x = 3sinx - {4sin}^{3}x}}$

So, using this, we get

$\sf \: = \: {(sin3x)}^{2}$

$\sf \: = \: {sin}^{2}3x$

Hence,

${\rm :\longmapsto\:\bf \: {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2} } = \bf{sin}^{2}3x$

Now, Consider, LHS

$\red{\rm :\longmapsto\:\sf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} + {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2}}$

$\rm \: = \: {cos}^{2}3x + {sin}^{2}3x$

$\rm \: = \: 1$

Hence,

$\red{\bf :\longmapsto\:\sf \: {cos}^{2}x {(3 - 4 {cos}^{2} x)}^{2} + {sin}^{2}x {(3 - 4 {sin}^{2} x)}^{2} = 1}$

Proved

Additional Information :-

$\boxed{ \bf{ \: sin2x = 2sinx \: cosx}}$

$\boxed{ \bf{ \: cos2x = 1 - {2sin}^{2}x}}$

$\boxed{ \bf{ \: cos2x = {2cos}^{2}x - 1}}$

$\boxed{ \bf{ \: cos2x = {cos}^{2}x - {sin}^{2} x}}$

$\boxed{ \bf{ \: 1 + cos2x = {2cos}^{2}x}}$

$\boxed{ \bf{ \: 1 - cos2x = {2sin}^{2}x}}$